Modular equation for a quotient of eta functions
Since I am not so much familiar with the theory of modular forms, I am offering a solution which uses theory of theta functions and their link with elliptic integrals. With this approach the functional relation of $y(\tau)$ is transformed into a modular equation of degree $3$ which can be independently verified using the theory of modular equations.
Let $\tau$ be a complex number with positive imaginary part so that $q = e^{2\pi i\tau}$ satisfies $|q| < 1$. The Dedekind $\eta$ function is defined as $$\eta(\tau) = e^{\pi i\tau/12}\prod_{n = 1}^{\infty}(1 - e^{2\pi in\tau}) = q^{1/24}\prod_{n = 1}^{\infty}(1 - q^{n}) = 2^{-1/6}\sqrt{\frac{2K}{\pi}}k^{1/12}k'^{1/3}\tag{1}$$ where $k, k', K$ associated with nome $q$. Further we have $$\eta(2\tau) = q^{1/12}\prod_{n = 1}^{\infty}(1 - q^{2n}) = 2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\tag{2}$$ Now let $l,l', L$ be associated with $q' = e^{6\pi i\tau} = q^{3}$. Then we have $$\eta(3\tau) = 2^{-1/6}\sqrt{\frac{2L}{\pi}}l^{1/12}l'^{1/3}\tag{3}$$ and $$\eta(6\tau) = 2^{-1/3}\sqrt{\frac{2L}{\pi}}(ll')^{1/6}\tag{4}$$ Beukers' paper talks about the function $$y(\tau) = \frac{\eta^{8}(6\tau)\eta^{4}(\tau)}{\eta^{8}(2\tau)\eta^{4}(3\tau)}\tag{5}$$ and mentions that it satisfies the following identity $$y(-1/6\tau) = \frac{y(\tau) - 1/9}{y(\tau) - 1}\tag{6}$$ Using the expression of $y(\tau)$ in terms of $\eta$ function and the functional equation $$\eta(-1/\tau) = \sqrt{-i\tau}\eta(\tau)\tag{7}$$ satisfied by the $\eta$ function we see that equation $(6)$ is equivalent to the following identity $$\frac{\eta^{8}(\tau)\eta^{4}(6\tau)}{\eta^{8}(3\tau)\eta^{4}(2\tau)} = \frac{9\eta^{8}(6\tau)\eta^{4}(\tau) - \eta^{8}(2\tau)\eta^{4}(3\tau)}{\eta^{8}(6\tau)\eta^{4}(\tau) - \eta^{8}(2\tau)\eta^{4}(3\tau)}\tag{8}$$ or $$\eta^{12}(\tau)\eta^{12}(6\tau) - \eta^{8}(\tau)\eta^{8}(2\tau)\eta^{4}(3\tau)\eta^{4}(6\tau) = 9\eta^{4}(\tau)\eta^{4}(2\tau)\eta^{8}(3\tau)\eta^{8}(6\tau) - \eta^{12}(2\tau)\eta^{12}(3\tau)$$ and using equations $(1)-(4)$ we can see that this is equivalent to $$\left(\frac{4KL}{\pi^{2}}\right)^{2}k'^{2}l - \left(\frac{2K}{\pi}\right)^{4}kk'^{2} = 9\left(\frac{2L}{\pi}\right)^{4}ll'^{2} - \left(\frac{4KL}{\pi^{2}}\right)^{2}kl'^{2}$$ Using $m = K/L$ for multiplier we see that the above is equivalent to $$\frac{k'^{2}kl}{m^{2}} - k^{2}k'^{2} = \frac{9kll'^{2}}{m^{4}} - \frac{k^{2}l'^{2}}{m^{2}}\tag{9}$$ Now we know from the theory of modular equations of degree $3$ that $$k^{2} = \frac{(m - 1)(3 + m)^{3}}{16m^{3}},\,l^{2} = \frac{(m - 1)^{3}(3 + m)}{16m}$$ and $$k'^{2} = \frac{(m + 1)(3 - m)^{3}}{16m^{3}},\,l'^{2} = \frac{(m + 1)^{3}(3 - m)}{16m}$$ so that $$kl = \frac{(m - 1)^{2}(3 + m)^{2}}{16m^{2}}$$ Putting the values of $k, l, k', l'$ in terms of $m$ we see (after some simple cancellations of factors on both sides of the equation $(9)$) that equation $(9)$ holds if we can establish $$(3 - m)^{2}(m - 1) - m(3 - m)^{2}(3 + m) = 9(m - 1)(m + 1)^{2} - m(3 + m)(m + 1)^{2}\tag{10}$$ Since each side is a polynomial of degree $4$ in $m$ it follows that the relation $(10)$ holds identically for all values of $m$ if it holds for at least $5$ distinct values of $m$. We can verify very easily that it holds for $m = 0, 1, -1, 3, -3$ and hence equation $(10)$ holds. Thus we have proved that $(9)$ also holds.
The approach given above is more of a verification of the identity $(6)$ which Beukers mentions in his paper. It is desirable to seek a proof based on some identity relating eta functions of argument $\tau, 2\tau, 3\tau$ or perhaps a proof based on theory of modular forms which shows directly that $y(-1/6\tau)$ is a rational function of $y(\tau)$.
There is a simple proof using two eta product identities. If $$ y_1(\tau) := \eta(2\tau)\eta^{9}(6\tau)\eta^{4}(\tau), \quad y_2(\tau) := \eta^{9}(2\tau)\eta^{4}(3\tau)\eta(6\tau), $$ then $\, y(\tau) = y_1(\tau)/y_2(\tau), \,$ and
$$ y_2(\tau) - 9y_1(\tau) = \eta(\tau)^9\eta(3\tau)\eta(6\tau)^4, \tag{1} $$
$$ y_2(\tau) - y_1(\tau) = \eta(\tau)\eta(2\tau)^4\eta(3\tau)^9, \tag{2} $$
$$ y_3(\tau) := \frac{y_2(\tau) - 9y_1(\tau)}{y_2(\tau) - y_1(\tau)} = 9\frac{y(\tau) - 1/9}{y(\tau) - 1}, \tag{3} $$
$$ y_3(\tau) = 9 \frac{\eta(\tau)^8 \eta(6\tau)^4}{\eta(2\tau)^4 \eta(3\tau)^8} = y\Big(\frac{-1}{6\tau}\Big). \tag{4}$$
Equation $(1)$ is eta product identity $t_{6,13,30}$ multiplied by $\,\eta(6\tau)\,$ and equation $(2)$ is eta product identity $t_{6,13,34}$ multiplied by $\,\eta(2\tau).\,$ Both identities have many different forms and references to proofs are listed in their entry in my Dedekind eta product identities website.